A force $\stackrel{}{}$\(\vec{F}=\alpha \hat{i}+3 \hat{j}+6 \hat{k}\) is acting at a point  \(\vec{r}=2 \hat{i}-6 \hat{j}-12 \hat{k}\). The value of \(\alpha\) for which angular momentum about the origin is conserved is:
1. \(-1\)
2. \(2\)
3. zero
4. \(1\)

A particle is executing a simple harmonic motion. Its maximum acceleration is \(\alpha\) and maximum velocity is \(\beta.\) Then its time period of vibration will be:
1. \(\dfrac {\beta^2}{\alpha^2}\)
2. \(\dfrac {\beta}{\alpha}\)
3. \(\dfrac {\beta^2}{\alpha}\)
4. \(\dfrac {2\pi \beta}{\alpha}\)

Light of wavelength \(500~\text{nm}\) is incident on metal with work function \(2.28~\text{eV}\). The de-Broglie wavelength of the emitted electron is:

At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen’s wavelet from the edge of the slit and the wavelet from the midpoint of the slit is:

On a frictionless surface, a block of mass \(M\) moving at speed \(v\) collides elastically with another block of the same mass \(M\) which is initially at rest. After the collision, the first block moves at an angle \(\theta\) to its initial direction and has a speed \(\frac{v}{3}\). The second block’s speed after the collision will be:

A potentiometer wire of length \(L\) and a resistance \(r\) are connected in series with a battery of EMF \(E_{0 }\) and resistance \(r_{1}\). An unknown EMF is balanced at a length l of the potentiometer wire. The EMF \(E\) will be given by:
1. \(\frac{L E_{0} r}{l r_{1}}\)
2. \(\frac{E_{0} r}{\left(\right. r + r_{1} \left.\right)} \cdot \frac{l}{L}\)
3. \(\frac{E_{0} l}{L}\)
4. \(\frac{L E_{0} r}{\left(\right. r + r_{1} \left.\right) l}\)

Young’s modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross-section, one of steel and another of brass, are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of:
1. \(1:2\)
2. \(2:1\)
3. \(4:1\)
4. \(1:1\)

The input signal given to a CE amplifier having a voltage gain of 150 is ${\mathrm{V}}_{\mathrm{i}}=2\cos \left[15\mathrm{t}+\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 3}}\right]$. The corresponding output signal will be:

1. $30<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{\mathrm{\pi }}{3}\right]$

2. $75<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{2\mathrm{\pi }}{3}\right]$

3. $2<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{5\mathrm{\pi }}{3}\right]$

4. $300<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{4\mathrm{\pi }}{3}\right]$

An automobile moves on a road with a speed of \(54~\text{kmh}^{-1}.\)  The radius of its wheels is \(0.45\) m and the moment of inertia of the wheel about its axis of rotation is \(3~\text{kg-m}^2.\) If the vehicle is brought to rest in \(15\) s, the magnitude of average torque transmitted by its brakes to the wheel is: